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Math Help - Inverse of a function

  1. #1
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    Inverse of a function

    f(x)=[x^2+5]/[x-3] where x not equal to 3
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  2. #2
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    Quote Originally Posted by write2rvr View Post
    f(x)=[x^2+5]/[x-3] where x not equal to 3
    set f(x) = y and make x the subject:

    yx - 3y = x^2 + 5

    x^2 - yx = -(5+3y)

    (x+\frac{y}{2})^2 - \frac{y^2}{4} = -5 - 3y <-- from completing the square

    (x+\frac{y}{2}) = \sqrt{\frac{y^2}{4} - 3y - 5}

    x = - \frac{y}{2} \sqrt{\frac{y^2}{4} - 3y - 5}

    Now set any y to x and set x to f^{-1}(x)

    f^{-1}(x) = - \frac{x}{2} \sqrt{\frac{x^2}{4} - 3x - 5}
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  3. #3
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    Quote Originally Posted by write2rvr View Post
    f(x)=[x^2+5]/[x-3] where x not equal to 3
    When you get to:

    . . . . .0 = x^2 - xy + 3y + 5

    ...you can always plug this into the Quadratic Formula, using a = 1, b = -y, and c = 3y + 5.
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